\newproblem{lay:5_2_20}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.2.20}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Use a property of determinants to show that $A$ and $A^T$ have the same characteristic polynomial.
}{
   % Solution
	The characteristic polynomial of $A$ is given by 
	\begin{center}
		$\det\{A-\lambda I\}$
	\end{center}
	For any matrix $X$, we know that $\det\{X\}=\det\{X^T\}$, then
	\begin{center}
		$\det\{A-\lambda I\}=\det\{(A-\lambda I)^T\}=\det\{A^T-\lambda I^T\}=\det\{A^T-\lambda I\}$
	\end{center}
	But this latter expression is the characteristic polynomial of $A^T$.
}
\useproblem{lay:5_2_20}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
